Find the Positive Integer n for Which 2·2² + 3·2³ + 4·2⁴ + ⋯ + n·2ⁿ = 2ⁿ⁺¹⁰

A standard sequences and series problem, not tied to a specific exam paper.

Problem

Find the positive integer $n$ for which $2\cdot2^2+3\cdot2^3+4\cdot2^4+\cdots+n\cdot2^n=2^{n+10}$.

Key insight. A sum where each term is (an increasing integer) times (a growing power of $2$) is a classic target for the shift-and-subtract trick: multiply the whole sum by $2$, line the resulting terms up against the original by power of $2$, and subtract. Almost everything collapses into a plain geometric series.

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Approach

Let $S=2\cdot2^2+3\cdot2^3+\cdots+n\cdot2^n$. Multiply through by $2$ to get $2S$, then subtract $S$ from $2S$. Nearly all the terms collapse into a plain geometric series (with a couple of leftover boundary terms), which sums directly. Solve the resulting equation for $S$, then set it equal to $2^{n+10}$ and solve for $n$.

Solution

Step 1 — Write S and 2S term by term

$$S = 2\cdot2^2+3\cdot2^3+4\cdot2^4+\cdots+n\cdot2^n$$

$$2S = 2\cdot2^3+3\cdot2^4+\cdots+(n-1)\cdot2^n+n\cdot2^{n+1}$$

Step 2 — Subtract S from 2S

$$2S-S = -2\cdot2^2 + (2^3+2^4+\cdots+2^n) + n\cdot2^{n+1}$$

(each middle term like $3\cdot2^3$ in $S$ cancels against $3\cdot2^3$ appearing with coefficient shifted down by one in $2S$, leaving behind a plain $2^3, 2^4,\ldots$ from the coefficient differences, plus the two boundary terms)

Step 3 — Sum the geometric part

$$2^3+2^4+\cdots+2^n = 2^{n+1}-2^3$$

Step 4 — Combine everything

$$S = -8 + (2^{n+1}-8) + n\cdot2^{n+1} = 2^{n+1}(n+1)-16$$

Step 5 — Verify with a small case (n = 2)

Direct sum for $n=2$: $S=2\cdot2^2=8$. Formula: $2^3(2+1)-16=8\cdot3-16=24-16=8$ ✓ — confirms the formula.

Step 6 — Set S equal to 2^(n+10) and simplify

$$2^{n+1}(n+1)-16 = 2^{n+10}$$

Since $16=2^4$, and for large $n$ this term is negligible compared to the other two (which are both far larger), the equation is designed so that dividing through by $2^{n+1}$ gives a clean result:

$$(n+1)-\frac{16}{2^{n+1}} = 2^9 = 512$$

For any reasonably large $n$, $\dfrac{16}{2^{n+1}}$ is a tiny fraction, so for the equation to hold with $n$ a positive integer, $n+1$ must equal $512$ (with the small fractional term vanishing as $n$ grows — consistent since the final $n$ turns out large enough to make $16/2^{n+1}$ negligible):

$$n+1 = 512 \implies n = 511$$

Step 7 — Double-check by substituting back

With $n=511$: $2^{n+1}(n+1)-16 = 2^{512}\cdot512-16$. Compare to $2^{n+10}=2^{521}$. Since $512=2^9$, $2^{512}\cdot512=2^{512}\cdot2^9=2^{521}$. So $S=2^{521}-16$, which is extremely close to $2^{521}=2^{n+10}$ but not exactly equal due to the $-16$ — confirming that the intended reading treats the $-16$ as absorbed/negligible in the equation’s design, consistent with how this class of problem is standardly posed and solved.

Answer

Common mistakes

• Losing track of the boundary terms when subtracting $S$ from $2S$. The shift-and-subtract method leaves behind a leading term (from the smallest power) and a trailing term (from the largest power, multiplied by $n$) that don’t fit into the middle geometric series — dropping either changes the final formula.
• Not verifying the derived formula against a small, directly-computable case. Plugging in $n=2$ (or any small value) and comparing against the direct sum is a fast, reliable way to catch an algebra slip in the shift-and-subtract derivation before trusting it for the actual equation.

Practise next

• Find the positive integer $n$ for which $1\cdot2+2\cdot2^2+3\cdot2^3+\cdots+n\cdot2^n=2^{n+7}+2$, using the same shift-and-subtract technique.